Question

4.8g of calcium is added to 3.6g of water. The following reaction occurs

Answer 1

Reaction of interest is: Ca + 2H2O ↔ Ca(OH)2 + H2

Answer 1:
Number of moles of calcium = $\frac{\text{weight of Ca (g)}}{\text{Atomic Weight}}$

Atomic weight of Ca = 40.078 g/mol

Therefore, Number of moles of calcium = $\frac{\text{4.8}}{\text{40.078}}$
                                                               = 0.1198.

Number of moles of Ca = 0.1198.
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Answer 2: 
Number of moles of water = $\frac{\text{weight of H2O (g)}}{\text{Molecular Weight}}$

Molecular weight of H2O = 18 g/mol

Therefore, Number of moles of water = $\frac{\text{3.6}}{\text{18}}$
                                                               = 0.2.

Number of moles of water = 0.2.
......................................................................................................................

Answer 3: 
Number of moles of water = 0.2
Number of moles of Ca = 0.1198

Here it can be seen that, Number of moles of water > Number of moles of Ca.
Hence, water is in excess. 
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Answer 4: 
Number of moles of water that is in excess = 0.2 - 0.1198
                                                                     = 0.0802

We know that, 1 mole of water ≡ 18 g of water
Therefore, 0.0802 mole of water ≡ 1.4436 g of water.

Thus, 1.4436 g of water is excess in the experiment.